Lesson 11: Limits and Continuity

Slide 1: Section 11.2 Limits and Continuity Math 21a February 29, 2008 Announcements Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b Office hours Tuesday, Wednesday 2–4pm SC 323 Midterm I Review Session 3/5, 6–7:30pm in SC Hall D Midterm I, 3/11, 7–9pm in SC Hall D Image: kaet44

Slide 2: Outline Introduction and definition Rules of limits Complications Showing a limit doesn’t exist Showing a limit does exist Continuity Worksheet

Slide 3: Where we’re going: derivatives of multivariable functions Recall that if f is a function that takes real numbers to real numbers, f (x + h) − f (x) f (x) = lim h→0 h

Slide 4: Where we’re going: derivatives of multivariable functions Recall that if f is a function that takes real numbers to real numbers, f (x + h) − f (x) f (x) = lim h→0 h We want to do the same thing in more than one variable. So we need to take limits in more than one dimension.

Slide 5: Definition We write lim f (x, y ) = L (x,y )→(a,b) and we say that the limit of f (x, y ) as (x, y ) approaches (a, b) is L if we can make the values of f (x, y ) as close to L as we like by taking the point (x, y ) to be sufficiently close to (a, b).

Slide 6: easy limits lim x =a (x,y )→(a,b) lim y =b (x,y )→(a,b) lim c =c (x,y )→(a,b)

Slide 7: Outline Introduction and definition Rules of limits Complications Showing a limit doesn’t exist Showing a limit does exist Continuity Worksheet

Slide 8: Like regular limits, limits of multivariable functions can be added subtracted multiplied composed divided, provided the limit of the denominator is not zero.

Slide 9: Limit of a Polynomial Example Find lim (x 5 + 4x 3 y − 5xy 2 ) (x,y )→(5,−2)

Slide 10: Limit of a Polynomial Example Find lim (x 5 + 4x 3 y − 5xy 2 ) (x,y )→(5,−2) Solution lim (x 5 + 4x 3 y − 5xy 2 ) = (5)5 + 4(5)3 (−2) − 5(5)(−2)2 (x,y )→(5,−2) = 3125 + 4(125)(−2) − 5(5)(4) = 2025.

Slide 11: Limit of a Rational Expression Example Compute x2 lim . (x,y )→(1,2) x 2 + y 2

Slide 12: Limit of a Rational Expression Example Compute x2 lim . (x,y )→(1,2) x 2 + y 2 Solution x2 (1)2 lim = (x,y )→(1,2) x 2 + y 2 (1)2 + (2)2 1 = 5

Slide 13: Outline Introduction and definition Rules of limits Complications Showing a limit doesn’t exist Showing a limit does exist Continuity Worksheet

Slide 14: The only real problem is a limit where the denominator goes to zero. If the numerator goes to some number and the denominator goes to zero then the quotient cannot have a limit.

Slide 15: The only real problem is a limit where the denominator goes to zero. If the numerator goes to some number and the denominator goes to zero then the quotient cannot have a limit. If on the other hand the numerator and denominator both go to zero we have no clue. Most “interesting” limits come from this. e.g., f (x + h) − f (x) f (x) = lim h→0 h

Slide 16: You probably remember this statement: Fact For a function f (x) of one variable, lim f (x) = L ⇐⇒ lim+ f (x) = L and lim f (x) = L x→a x→a x→a−

Slide 17: You probably remember this statement: Fact For a function f (x) of one variable, lim f (x) = L ⇐⇒ lim+ f (x) = L and lim f (x) = L x→a x→a x→a− For functions of two variables, “left-hand limits” and “right-hand limits” aren’t enough.

Slide 18: Showing a limit doesn’t exist Theorem Suppose lim f (x, y ) = L. Then the limit of f as (x,y )→(a,b) (x, y ) → (a, b) is L along all paths through (a, b).

Slide 19: Showing a limit doesn’t exist Theorem Suppose lim f (x, y ) = L. Then the limit of f as (x,y )→(a,b) (x, y ) → (a, b) is L along all paths through (a, b). There are two contrapositives to this statement:

Slide 20: Showing a limit doesn’t exist Theorem Suppose lim f (x, y ) = L. Then the limit of f as (x,y )→(a,b) (x, y ) → (a, b) is L along all paths through (a, b). There are two contrapositives to this statement: If there is a path through (a, b) along which the limit does not exist, the two-dimensional limit does not exist

Slide 21: Showing a limit doesn’t exist Theorem Suppose lim f (x, y ) = L. Then the limit of f as (x,y )→(a,b) (x, y ) → (a, b) is L along all paths through (a, b). There are two contrapositives to this statement: If there is a path through (a, b) along which the limit does not exist, the two-dimensional limit does not exist If there are two paths through (a, b) along which the limits exist but disagree, the two-dimensional limit does not exist

Slide 22: Example x Show lim does not exist. (x,y )→(0,0) x 2 + y2

Slide 23: Example x Show lim does not exist. (x,y )→(0,0) x 2 + y2 Solution Follow a path towards (0, 0) along the line y = 0. x 1 lim f (x, 0) = lim = lim x→0 x→0 x2 +0 2 x→0 x x which does not exist. So lim does not exist. (x,y )→(0,0) x 2 + y2

Slide 24: Example x Show lim does not exist. (x,y )→(0,0) x 2 + y2 Solution Follow a path towards (0, 0) along the line y = 0. x 1 lim f (x, 0) = lim = lim x→0 x→0 x2 +0 2 x→0 x x which does not exist. So lim does not exist. (x,y )→(0,0) x 2 + y2

Slide 25: We can see the problems in a graph. 2 2 x 1 1 2 + y2 = c ⇐⇒ x− + y2 = x c 2c

Slide 26: We can see the problems in a graph. 2 2 x 1 1 2 + y2 = c ⇐⇒ x− + y2 = x c 2c 1 0.5 0 -0.5 -1 -1 -0.5 0 0.5 1

Slide 27: We can see the problems in a graph. 2 2 x 1 1 2 + y2 = c ⇐⇒ x− + y2 = x c 2c 1 0.5 0 10 5 1 0 -5 0.5 -0.5 -10 -1 0 -0.5 0 -0.5 -1 0.5 -1 -0.5 0 0.5 1 1 -1

Slide 28: Example x2 Show lim does not exist. (x,y )→(0,0) x 2 + y 2

Slide 29: Example x2 Show lim does not exist. (x,y )→(0,0) x 2 + y 2 Solution Follow a path towards (0, 0) along the line y = 0:

Slide 30: Example x2 Show lim does not exist. (x,y )→(0,0) x 2 + y 2 Solution Follow a path towards (0, 0) along the line y = 0: x2 lim f (x, 0) = lim = lim 1 = 1 x→0 x→0 x 2 + 02 x→0

Slide 31: Example x2 Show lim does not exist. (x,y )→(0,0) x 2 + y 2 Solution Follow a path towards (0, 0) along the line y = 0: x2 lim f (x, 0) = lim = lim 1 = 1 x→0 x→0 x 2 + 02 x→0 Now follow a path towards (0, 0) along the line x = 0:

Slide 32: Example x2 Show lim does not exist. (x,y )→(0,0) x 2 + y 2 Solution Follow a path towards (0, 0) along the line y = 0: x2 lim f (x, 0) = lim = lim 1 = 1 x→0 x→0 x 2 + 02 x→0 Now follow a path towards (0, 0) along the line x = 0: 02 lim f (0, y ) = lim = lim 0 = 0 y →0 x→0 02 + y 2 x→0 So the limit as (x, y ) → (0, 0) cannot exist.

Slide 33: Again, we can see the problems in a graph. x2 1−c = c ⇐⇒ y = ± x x2 + y2 c

Slide 34: Again, we can see the problems in a graph. x2 1−c = c ⇐⇒ y = ± x x2 + y2 c 1 0.5 0 -0.5 -1 -1 -0.5 0 0.5 1

Slide 35: Again, we can see the problems in a graph. x2 1−c = c ⇐⇒ y = ± x x2 + y2 c 1 0.5 0 1 0.75 1 0.5 0.25 0.5 -0.5 0 -1 0 -0.5 0 -0.5 -1 0.5 -1 -0.5 0 0.5 1 1 -1

Slide 36: Showing a limit does exist This is often harder. No single method always works. Example Show x3 lim = 0. (x,y )→(0,0) x 2 + y 2

Slide 37: Showing a limit does exist This is often harder. No single method always works. Example Show x3 lim = 0. (x,y )→(0,0) x 2 + y 2 Solution From the last problem we know that x2 0≤ ≤1 x2 + y2 for all x and y , not both 0.

Slide 38: Showing a limit does exist This is often harder. No single method always works. Example Show x3 lim = 0. (x,y )→(0,0) x 2 + y 2 Solution From the last problem we know that x2 0≤ ≤1 x2 + y2 for all x and y , not both 0. So if x > 0, x3 0≤ ≤ x. x2 + y2

Slide 39: Showing a limit does exist This is often harder. No single method always works. Example Show x3 lim = 0. (x,y )→(0,0) x 2 + y 2 Solution From the last problem we know that x2 0≤ ≤1 x2 + y2 for all x and y , not both 0. So if x > 0, x3 0≤ ≤ x. x2 + y2 As x → 0+ , the fraction in the middle must go to 0!

Slide 40: Another way Switch to polar coordinates! x3 (r cos θ)3 lim = lim (x,y )→(0,0) x 2 + y 2 (r ,θ)→(0,0) (r cos θ)2 + (r sin θ)2 r 3 cos3 θ = lim (r ,θ)→(0,0) r 2 cos2 θ + r 2 sin2 θ = lim r cos3 θ = 0 · 1 = 0. (r ,θ)→(0,0)

Slide 41: Outline Introduction and definition Rules of limits Complications Showing a limit doesn’t exist Showing a limit does exist Continuity Worksheet

Slide 42: Continuity Definition A function f of two variables is called continuous at (a, b) if lim f (x, y ) = f (a, b). (x,y )→(a,b) We say f is continuous on D if f is continuous at every point (a, b) in D.

Slide 43: Outline Introduction and definition Rules of limits Complications Showing a limit doesn’t exist Showing a limit does exist Continuity Worksheet

Slide 44: Worksheet Image: Erick Cifuentes